NEET/JEE Coaching Scholarship
ApplyGet up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
A whole number starts with 0 and ends at infinity. And a number with 2 digits starts at 10 and ends at 99. In this article, we will figure out the 2-digit numbers divisible by 7. We will first understand the formula which helps us to find the solution.
To find out n-digit numbers that are divisible by m, we use the following steps:
Write the numbers in arithmetic progression
Find the number of terms in the arithmetic progression
Now, apply these steps to find 2-digit numbers divisible by 7
The first 2-digit number divisible by 7 = 14
The last 2-digit number divisible by 7 = 98
Hence, the First term is 14, and the common difference, d is 7
To find the number of terms in the above arithmetic progression, we use the formula for the nth term of an AP which is given by:
\begin{equation}
a_n=a+(n-1) d
\end{equation}
Here, an = 98
a = 14
d= 7
And, We have to find the value of n.
Now, putting the values, we get:
\begin{equation}
\begin{aligned}
&98=14+(n-1) 7 \
&84=7 n-7 \
&7 n=91 \
&n=13
\end{aligned}
\end{equation}
Therefore, 13 numbers with 2-digits are divisible by 7.
Below are the 13 numbers by 7
14 | 42 | 70 | 98 |
21 | 49 | 77 | |
28 | 56 | 84 | |
35 | 63 | 91 |
Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide